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Question Number 143588    Answers: 3   Comments: 0

Question Number 143582    Answers: 0   Comments: 0

x^3 −x−c=0 let x=t+h (t−s){t^3 +3ht^2 +(3h^2 −1)t +(h^3 −h−c)}=0 let t=p+q (p+q−s){p^3 +q^3 +3pq(p+q) +3h(p^2 +q^2 )+6hpq +(3h^2 −1)(p+q)+(h^3 −h−c)} = 0 p^4 +q^4 +pq(p^2 +q^2 )+3pq(p+q)^2 +3h(p^2 +q^2 )(p+q)+6hpq(p+q) +(3h^2 −1)(p+q)^2 +(h^3 −h−c)(p+q) −s(p^3 +q^3 )−3pqs(p+q) −3hs(p^2 +q^2 )−6hspq +(3h^2 −1)s(p+q) −s(h^3 −h−c)=0 let p^4 +q^4 =p^2 +q^2 (p^2 +q^2 )^2 −(p^2 +q^2 )−2p^2 q^2 =0 ⇒ p^2 +q^2 =(1/2)±(√((1/4)+2p^2 q^2 )) say z=(1/2)+(√((1/4)+2m^2 )) (p+q)^2 =z+2m ⇒z(1+m+3m+3h^2 −1−3hs) +(z+2m)^(1/2) {3hz+6mh +h^3 −h−c−3ms −s(z−m)+(3h^2 −1)s} +6m^2 +2(3h^2 −1)m−6hsm −s(h^3 −h−c)=0 let (3h−s)z+m(6h−2s) +h^3 −h−c+3sh^2 −s=0 & 4m+3h(h−s)=0 ⇒ 2(3h−s)z−3h(h−s)(3h−s) +3sh^2 −s+h^3 −h−c=0 let s=−3h ⇒ 12hz−80h^3 −h−c=0 & m=−3h^2 ⇒ 54h^4 −6h^2 (3h^2 −1) −54h^4 +3h(h^3 −h−c)=0 ⇒ 15h^4 −3h^2 +3ch=0 And if h≠0 ⇒ 5h^3 −h+c=0 D=(c^2 /(100))−((1/(15)))^3 ⇒ D≥0 if c≥ (2/(3(√(15)))) m=−3h^2 ; s=−3h; And 12hz−80h^3 −h−c=0 ⇒ z=((20h^2 )/3)+(1/(12))+(c/(12h)) p+q=(√(z+2m)) =(√(((2h^2 )/3)+((c+h)/(12h)))) x=t+h=p+q+h x=h+(√(((2h^2 )/3)+((c+h)/(12h)))) but (c/(12h))>−(((2h^2 )/3)+(1/(12))) ⇒ c<−8h^3 −h ⇒ 8h^3 +h+c<0 5h^3 −h+c=0 ⇒ 13h^3 +2c<0 ......

$${x}^{\mathrm{3}} −{x}−{c}=\mathrm{0} \\ $$$${let}\:\:{x}={t}+{h} \\ $$$$\left({t}−{s}\right)\left\{{t}^{\mathrm{3}} +\mathrm{3}{ht}^{\mathrm{2}} +\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right){t}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({h}^{\mathrm{3}} −{h}−{c}\right)\right\}=\mathrm{0} \\ $$$${let}\:{t}={p}+{q} \\ $$$$\left({p}+{q}−{s}\right)\left\{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} +\mathrm{3}{pq}\left({p}+{q}\right)\right. \\ $$$$\:\:+\mathrm{3}{h}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)+\mathrm{6}{hpq} \\ $$$$\left.\:\:+\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)\left({p}+{q}\right)+\left({h}^{\mathrm{3}} −{h}−{c}\right)\right\} \\ $$$$\:\:=\:\mathrm{0} \\ $$$${p}^{\mathrm{4}} +{q}^{\mathrm{4}} +{pq}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)+\mathrm{3}{pq}\left({p}+{q}\right)^{\mathrm{2}} \\ $$$$+\mathrm{3}{h}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)\left({p}+{q}\right)+\mathrm{6}{hpq}\left({p}+{q}\right) \\ $$$$+\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)\left({p}+{q}\right)^{\mathrm{2}} \\ $$$$+\left({h}^{\mathrm{3}} −{h}−{c}\right)\left({p}+{q}\right) \\ $$$$−{s}\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right)−\mathrm{3}{pqs}\left({p}+{q}\right) \\ $$$$−\mathrm{3}{hs}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)−\mathrm{6}{hspq} \\ $$$$+\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right){s}\left({p}+{q}\right) \\ $$$$−{s}\left({h}^{\mathrm{3}} −{h}−{c}\right)=\mathrm{0} \\ $$$${let}\:\:{p}^{\mathrm{4}} +{q}^{\mathrm{4}} ={p}^{\mathrm{2}} +{q}^{\mathrm{2}} \\ $$$$\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)^{\mathrm{2}} −\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)−\mathrm{2}{p}^{\mathrm{2}} {q}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:{p}^{\mathrm{2}} +{q}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}{p}^{\mathrm{2}} {q}^{\mathrm{2}} } \\ $$$${say}\:\:\:\:{z}=\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}{m}^{\mathrm{2}} } \\ $$$$\left({p}+{q}\right)^{\mathrm{2}} ={z}+\mathrm{2}{m} \\ $$$$\Rightarrow{z}\left(\mathrm{1}+{m}+\mathrm{3}{m}+\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}−\mathrm{3}{hs}\right) \\ $$$$+\left({z}+\mathrm{2}{m}\right)^{\mathrm{1}/\mathrm{2}} \left\{\mathrm{3}{hz}+\mathrm{6}{mh}\right. \\ $$$$\:\:\:\:+{h}^{\mathrm{3}} −{h}−{c}−\mathrm{3}{ms} \\ $$$$\left.\:\:\:\:−{s}\left({z}−{m}\right)+\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right){s}\right\} \\ $$$$+\mathrm{6}{m}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right){m}−\mathrm{6}{hsm} \\ $$$$−{s}\left({h}^{\mathrm{3}} −{h}−{c}\right)=\mathrm{0} \\ $$$${let}\:\:\:\left(\mathrm{3}{h}−{s}\right){z}+{m}\left(\mathrm{6}{h}−\mathrm{2}{s}\right) \\ $$$$\:\:\:\:\:\:\:+{h}^{\mathrm{3}} −{h}−{c}+\mathrm{3}{sh}^{\mathrm{2}} −{s}=\mathrm{0} \\ $$$$\&\:\:\mathrm{4}{m}+\mathrm{3}{h}\left({h}−{s}\right)=\mathrm{0}\:\:\Rightarrow \\ $$$$\:\mathrm{2}\left(\mathrm{3}{h}−{s}\right){z}−\mathrm{3}{h}\left({h}−{s}\right)\left(\mathrm{3}{h}−{s}\right) \\ $$$$\:\:+\mathrm{3}{sh}^{\mathrm{2}} −{s}+{h}^{\mathrm{3}} −{h}−{c}=\mathrm{0} \\ $$$${let}\:\:{s}=−\mathrm{3}{h} \\ $$$$\Rightarrow\:\mathrm{12}{hz}−\mathrm{80}{h}^{\mathrm{3}} −{h}−{c}=\mathrm{0} \\ $$$$\&\:\:{m}=−\mathrm{3}{h}^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{54}{h}^{\mathrm{4}} −\mathrm{6}{h}^{\mathrm{2}} \left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\:\:\:\:\:−\mathrm{54}{h}^{\mathrm{4}} +\mathrm{3}{h}\left({h}^{\mathrm{3}} −{h}−{c}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{15}{h}^{\mathrm{4}} −\mathrm{3}{h}^{\mathrm{2}} +\mathrm{3}{ch}=\mathrm{0} \\ $$$${And}\:{if}\:{h}\neq\mathrm{0}\:\:\Rightarrow \\ $$$$\:\:\mathrm{5}{h}^{\mathrm{3}} −{h}+{c}=\mathrm{0} \\ $$$${D}=\frac{{c}^{\mathrm{2}} }{\mathrm{100}}−\left(\frac{\mathrm{1}}{\mathrm{15}}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\:\:{D}\geqslant\mathrm{0}\:\:{if}\:{c}\geqslant\:\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{15}}} \\ $$$${m}=−\mathrm{3}{h}^{\mathrm{2}} \:;\:{s}=−\mathrm{3}{h};\:\:{And} \\ $$$$\:\mathrm{12}{hz}−\mathrm{80}{h}^{\mathrm{3}} −{h}−{c}=\mathrm{0} \\ $$$$\Rightarrow\:\:{z}=\frac{\mathrm{20}{h}^{\mathrm{2}} }{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{12}}+\frac{{c}}{\mathrm{12}{h}} \\ $$$${p}+{q}=\sqrt{{z}+\mathrm{2}{m}} \\ $$$$\:\:\:\:\:\:\:\:=\sqrt{\frac{\mathrm{2}{h}^{\mathrm{2}} }{\mathrm{3}}+\frac{{c}+{h}}{\mathrm{12}{h}}} \\ $$$${x}={t}+{h}={p}+{q}+{h} \\ $$$$\:\:\:{x}={h}+\sqrt{\frac{\mathrm{2}{h}^{\mathrm{2}} }{\mathrm{3}}+\frac{{c}+{h}}{\mathrm{12}{h}}} \\ $$$${but}\:\:\:\frac{{c}}{\mathrm{12}{h}}>−\left(\frac{\mathrm{2}{h}^{\mathrm{2}} }{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{12}}\right) \\ $$$$\Rightarrow\:\:{c}<−\mathrm{8}{h}^{\mathrm{3}} −{h} \\ $$$$\Rightarrow\:\:\mathrm{8}{h}^{\mathrm{3}} +{h}+{c}<\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{5}{h}^{\mathrm{3}} −{h}+{c}=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{13}{h}^{\mathrm{3}} +\mathrm{2}{c}<\mathrm{0} \\ $$$$...... \\ $$

Question Number 143578    Answers: 0   Comments: 2

Question Number 143576    Answers: 0   Comments: 0

find L(((arctanx)/x))

$${find}\:{L}\left(\frac{{arctanx}}{{x}}\right) \\ $$

Question Number 143575    Answers: 1   Comments: 0

find L(e^(−(√x)) )

$${find}\:{L}\left({e}^{−\sqrt{{x}}} \right) \\ $$

Question Number 143570    Answers: 1   Comments: 0

Question Number 143715    Answers: 0   Comments: 0

∫_0 ^(π/4) tanx∙Li(tan^2 x)dx

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{tanx}\centerdot\mathrm{Li}\left(\mathrm{tan}^{\mathrm{2}} \mathrm{x}\right)\mathrm{dx} \\ $$

Question Number 143562    Answers: 0   Comments: 0

Question Number 143561    Answers: 3   Comments: 0

lim_(x→∞) (1−(2/x^ )+(1/x^2 ))^x =? lim_(n→∞) ((√n)−(√(n−1)))(√(n+1)) =?

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}−\frac{\mathrm{2}}{{x}^{} }+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{{x}} =? \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\sqrt{{n}}−\sqrt{{n}−\mathrm{1}}\right)\sqrt{{n}+\mathrm{1}}\:=? \\ $$

Question Number 143556    Answers: 0   Comments: 0

∫_0 ^∞ (dx/(x^α (lnx)^β ))

$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\mathrm{x}^{\alpha} \left(\mathrm{lnx}\right)^{\beta} } \\ $$

Question Number 143546    Answers: 1   Comments: 0

calculate ∫_0 ^∞ ((log^2 x)/((8+x^4 )^2 ))dx

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{log}^{\mathrm{2}} \mathrm{x}}{\left(\mathrm{8}+\mathrm{x}^{\mathrm{4}} \right)^{\mathrm{2}} }\mathrm{dx} \\ $$

Question Number 143542    Answers: 0   Comments: 0

calculate the polar integral that give the area of the region bunded by the curves r=2 ,r=4cosθ and ,r cosθ=3

$${calculate}\:{the}\:{polar}\:{integral}\:{that} \\ $$$$\:{give}\:{the}\:{area}\:{of}\:{the}\:{region}\:{bunded}\:{by}\:{the}\:{curves}\: \\ $$$$ \\ $$$${r}=\mathrm{2}\:,{r}=\mathrm{4}{cos}\theta\:{and}\:,{r}\:{cos}\theta=\mathrm{3}\: \\ $$$$ \\ $$

Question Number 143531    Answers: 1   Comments: 0

lim_(x→+∞) ((2(√x)+3 (x)^(1/3) +5 (x)^(1/5) )/( (√(3x−2)) +((2x−3))^(1/3) )) =?

$$\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\frac{\mathrm{2}\sqrt{{x}}+\mathrm{3}\:\sqrt[{\mathrm{3}}]{{x}}\:+\mathrm{5}\:\sqrt[{\mathrm{5}}]{{x}}}{\:\sqrt{\mathrm{3}{x}−\mathrm{2}}\:+\sqrt[{\mathrm{3}}]{\mathrm{2}{x}−\mathrm{3}}}\:=? \\ $$

Question Number 143532    Answers: 1   Comments: 0

lim_(x→0) ((1−(√(1+x^2 )) cos x)/(tan^4 x)) =?

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\mathrm{cos}\:{x}}{\mathrm{tan}\:^{\mathrm{4}} {x}}\:=? \\ $$

Question Number 143523    Answers: 1   Comments: 0

Determiner l′origine de laplace 1−F(p)=(p/((p+2)^2 ))

$${Determiner}\:{l}'{origine}\:{de}\:{laplace} \\ $$$$\mathrm{1}−{F}\left({p}\right)=\frac{{p}}{\left({p}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$

Question Number 143521    Answers: 0   Comments: 1

Question Number 143519    Answers: 3   Comments: 0

The first two terms of the {a_n } series are defind as a_n =a_(n−1) +a_(n−2) for the general term a_1 =5, a_2 =8 and n≥3 . since the L=lim_(n→∞) (a_(n+1) /a_n ) what is the value of L

$$ \\ $$$${The}\:{first}\:{two}\:{terms}\:{of}\:{the}\:\left\{{a}_{{n}} \right\}\:{series}\:\:\:{are}\:{defind}\:{as}\:{a}_{{n}} ={a}_{{n}−\mathrm{1}} +{a}_{{n}−\mathrm{2}} \:\:{for}\:{the}\:{general}\:{term} \\ $$$$\:{a}_{\mathrm{1}} =\mathrm{5},\:{a}_{\mathrm{2}} =\mathrm{8}\:{and}\:{n}\geqslant\mathrm{3}\:. \\ $$$${since}\:{the}\:{L}={li}\underset{{n}\rightarrow\infty} {{m}}\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }\:\:{what}\:{is}\:{the}\:{value}\:{of}\:{L} \\ $$$$ \\ $$

Question Number 143516    Answers: 2   Comments: 1

Question Number 143508    Answers: 3   Comments: 0

..........Calculus........ i: 𝛗_1 :=∫_0 ^( 1) ((ln^2 (1−x).ln(x))/x)dx ii: 𝛗_2 := ∫_0 ^( 1) ((ln^2 (x).ln(1−x))/x) dx iii : 𝛗_3 :=∫_0 ^( 1) ((ln^2 (x).ln(1+x))/x)dx

$$ \\ $$$$\:\:\:\:\:\:\:\:..........{Calculus}........ \\ $$$$\:\:\:\:{i}:\:\:\:\boldsymbol{\phi}_{\mathrm{1}} :=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right).{ln}\left({x}\right)}{{x}}{dx} \\ $$$$\:\:\:{ii}:\:\:\:\boldsymbol{\phi}_{\mathrm{2}} :=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({x}\right).{ln}\left(\mathrm{1}−{x}\right)}{{x}}\:{dx} \\ $$$$\:\:{iii}\::\:\boldsymbol{\phi}_{\mathrm{3}} \::=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({x}\right).{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx} \\ $$

Question Number 143507    Answers: 1   Comments: 0

10(√(20))+13(√(45))

$$\mathrm{10}\sqrt{\mathrm{20}}+\mathrm{13}\sqrt{\mathrm{45}} \\ $$

Question Number 143506    Answers: 1   Comments: 0

∫_1 ^∞ (1/(e^(−x) +e^x )) dx=?

$$\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{\mathrm{1}}{{e}^{−{x}} +{e}^{{x}} }\:{dx}=? \\ $$$$ \\ $$

Question Number 143505    Answers: 1   Comments: 1

Question Number 143501    Answers: 0   Comments: 0

Question Number 143502    Answers: 0   Comments: 0

Question Number 143499    Answers: 1   Comments: 1

Question Number 143514    Answers: 0   Comments: 1

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